# Self-linkage of the orthogonal group $O_n({\mathbb R})$.

In Exercise 153 of my list, it is proved that the connected components $SO_2({\mathbb R})$ and $O_2^-({\mathbb R})$ of the orthogonal group are linked as curves in the three-dimensional sphere defined by $\|M\|=1$ in the operator norm.

Question. Consider the case of $n\times n$ matrices. The real orthogonal group has two connected components $SO_n({\mathbb R})$ and $O_n^-({\mathbb R})$, each one a submanifold of dimension $n(n-1)/2$. Let $s_1(M)\ge\cdots\ge s_n(M)(\ge0)$ be the singular values of a matrix $M$. We have $s_1(M)=\|M\|$. Define the set $V_n$ by the identities $s_1(M)=\cdots=s_{n-1}(M)=1$. Is it a submanifold of dimension $n^2-n+1$ ? In $V_n$, are $SO_n({\mathbb R})$ and $O_n^-({\mathbb R})$ linked ?

Nota. The exercise in $M_2({\mathbb R})$ is included in the forthcoming second edition of my book Matrices'' as Exercise 21 in chapter 10.

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The dimension of $V_n$ is at most $\binom{n}{2} +n$. For $n>2$, this is less than $2 \binom{n}{2} +1$.

It is convenient to solve a more general problem. Fix $s_1 \geq s_2 \geq \cdots \geq s_n \geq 0$. Let $M(s_1, \ldots, s_n)$ be the space of matrices with singular values $s_i$. Let $d_1$, $d_2$, ..., $d_k$ be the list of multiplicites with which the $s$'s occur, so $\sum d_i = n$. For example, if the $s$'s are $(1,1,1,\ldots, 1, r)$ then $(d_1, d_2) = (n-1, 1)$.

I claim that $\dim M(s) = 2 \binom{n}{2} - \sum \binom{d_i}{2}$. Proof: The group $O(n) \times O(n)$ acts transitively on $M(s)$, so the dimension of $M(s)$ is $\dim ( O(n) \times O(n)) = 2 \binom{n}{2}$ minus the dimensional of the stabilizer of the diagonal matrix $d(s) := \mathrm{diag}(s_1, s_2, \ldots, s_{n-1}, s_n)$. We compute the Lie algebra of the stabilizer. Let $g$ and $h$ be skew-symmetric matrices. Then, to first order, $e^g d(s) e^h = d(s)$ if and only if $g d(s) + d(s) h=0$.

For every $(i,j)$, this gives the equations $g_{ij} s_j + s_i h_{ij}$ and $s_i g_{ij} + s_j h_{ij} =0$. If $s_i \neq \pm s_j$, this forces $g_{ij} = h_{ij}=0$. If $s_i = \pm s_j$ then the sign must be $+$, as the $s_i$ are nonnegative, and we get $g_{ij} = - h_{ij}$. So the Lie algebra of the stabilizer has a basis element for every $(i,j)$ with $s_i = s_j$, and we see that the dimension of the stabilizer is $\sum \binom{d_i}{2}$.

You are interested in $\bigcup_{0 \leq r \leq 1} M(1,1,\ldots,1,r)$. Each space in the union has dimension $2\binom{n}{2} - \binom{n-1}{2} = \binom{n}{2} + n-1$, except for the $r=1$ term which is even smaller. Thus, the union has dimension at most $\binom{n}{2} +n$. I have not been able to work out whether the union is a manifold near the boundary points $r=0$ and $r=1$.

One way to salvage your question would be to look at matrices whose singular values are $(1,1-r/(n-1), 1-2r/(n-1), \ldots, 1-(n-2)r/(n-1), 1-r)$, for $r \in [0,1]$.

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Thank you ! I agree with your calculation and your suggestion. After all, it was important to impose $s_1=1$, but for the choice of $s_2,\ldots,s_{n-1}$, I was lazy. I change my question accordingly. – Denis Serre Oct 14 '10 at 13:36
@David. Actually, the choice of smaller $s_2,\ldots$ does not work, because the corresponding set does not contain $O_n({\mathbb R})$. Orthogonal matrices have $s_1=\cdots=s_n=1$, and are characterized by this property. I'll think to how fix the question. – Denis Serre Oct 14 '10 at 13:43
But when $r=0$, we have $1-kr/(n-1)=1$. Doesn't that fix the question? – David Speyer Oct 14 '10 at 13:48
At first glance, the question makes sense if $n=1+\binom{m+1}{2}$ (that is $n=2,4,7,11,16,...$) and $V_n$ is defined by $s_1(M)=\cdots=s_m(M)=1$. – Denis Serre Oct 14 '10 at 13:57
That also works. – David Speyer Oct 14 '10 at 14:23